70638637 control system lab manual opt

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DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

EE2257 CONTROL SYSTEMS LABORATORY MANUAL

NAME :

CLASS :

SEMESTER :

ROLL NUMBER :

REGISTER NUMBER :

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INDEX

S.No.

Date Title of Experiment PageNo.

Marks Signature





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SYLLABUS

EE2257 CONTROL SYSTEM LABORATORY 0 0 3 2

1. Determination of transfer function of DC Servomotor2. Determination of transfer function of AC Servomotor.

3. Analog simulation of Type - 0 and Type – 1 systems4. Determination of transfer function of DC Generator

5. Determination of transfer function of DC Motor6. Stability analysis of linear systems

7. DC and AC position control systems8. Stepper motor control system

9. Digital simulation of first order systems10. Digital simulation of second order systems

P = 45 Total = 45DETAILED SYLLABUS

1. Determination of Transfer Function Parameters of a DC ServoMotor

Aim

To derive the transfer function of the given D.C Servomotor and experimentallydetermine the transfer function parameters

Exercise

1. Derive the transfer function from basic principles for a separately excited DC

motor.2. Determine the armature and field parameters by conducting suitable experiments.3. Determine the mechanical parameter by conducting suitable experiments.

4. Plot the frequency response.

Equipment

1. DC servo motor : field separately excited – loading facility – variable voltagesource - 1 No

2. Tachometer : 1 No3. Multimeter : 2 Nos

4. Stop watch : 1 No

2. Determination of Transfer Function Parameters of AC Servo Motor

Aim

To derive the transfer function of the given A.C Servo Motor and experimentallydetermine the transfer function parameters





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Exercise

1. Derive the transfer function of the AC Servo Motor from basic Principles.2. Obtain the D.C gain by operating at rated speed.

3. Determine the time constant (mechanical)

4. Plot the frequency response

Equipment1. AC Servo Motor : Minimum of 100w – necessary sources for main winding and

control winding – 1 No2. Tachometer : 1 No3. Stopwatch : 1 No

4. Voltmeter : 1 No

3. Analog Simulation of Type-0 And Type-1 System

AimTo simulate the time response characteristics of I order and II order, type 0 and type-1systems.

Exercise1. Obtain the time response characteristics of type – 0 and type-1, I order and II

order systems mathematically.2. Simulate practically the time response characteristics using analog rigged up

modules.3. Identify the real time system with similar characteristics.

Equipment 1. Rigged up models of type-0 and type-1 system using analog components.

2. Variable frequency square wave generator and a normal CRO - 1 No(or)

DC source and storage Oscilloscope - 1 No

4. Determination of Transfer function of DC Generator

Aim

To determine the transfer function of DC generator

Exercise1. Obtain the transfer function of DC generator by calculating and gain

Equipment 1. DC Generator

2. Tachometer3. Various meters

4. Stop watch





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5. Determination of Transfer function of DC Motor

Aim

To determine the transfer function of DC motor

Exercise

1. Obtain the transfer function of DC motor by calculating and gain

Equipment 1. DC Motor

2. Tachometer3. Various meters

4. Stop watch

6. Stability Analysis of Linear Systems

Aim

To analyse the stability of linear systems using Bode / Root locus / Nyquist plot

Exercise

1. Write a program to obtain the Bode plot / Root locus / Nyquist plot for the givensystem

2. Access the stability of the given system using the plots obtained3. Compare the usage of various plots in assessing stability

Equipment

1. System with MATLAB / MATHCAD / equivalent software - 3 user license

7. DC and AC position Control Systems

AimTo study the AC and DC position control system and draw the error characteristics

between set point and error.

Exercise

1. To study various position control systems and calculate the error between set pointand output position

2. To measure outputs at various points (between stages)

Equipment

1. AC and DC position control kit with DC servo motor.2. Power transistor

3. Adder





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8. Stepper Motor Control System

Aim

To study the working of stepper motor

Exercise1. To verify the working of the stepper motor rotation using microprocessor.

Equipment

1. Stepping motor2. Microprocessor kit3. Interfacing card

4. Power supply

9. Digital Simulation of First Order System

AimTo digitally simulate the time response characteristics of first -order system

Exercise1. Write a program or build the block diagram model using the given software.

2. Obtain the impulse, step and sinusoidal response characteristics.3. Identify real time systems with similar characteristics.

Equipment1. System with MATLAB / MATHCAD (or) equivalent software - minimum 3

user license.

10. Digital Simulation of Second Order Systems

Aim

To digitally simulate the time response characteristics of second -order system

Exercise

1. Write a program or build the block diagram model using the givensoftware.

2. Obtain the impulse, step and sinusoidal response characteristics.3. Identify real time systems with similar characteristics.

EquipmentSystem with MATLAB / MATHCAD (or) equivalent software - minimum 3 user

license.





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LIST OF EXPERIMENTS

FIRST CYCLE:

1. Determination of transfer function of armature controlled DC

servomotor.

2. Determination of transfer function of field controlled DC servomotor.

3. Determination of transfer function of AC servomotor.

4.

Determination of transfer function of separately excited DC generator.

5. Determination of transfer function of DC motor.

6. DC position control system.

SECOND CYCLE:

7. Analog simulation of Type-0 and Type-1 systems.

8. Digital simulation of first order systems.

9. Digital simulation of second order systems

10. Stability analysis of linear systems.

11. Stepper motor control system.

12. AC position control system.





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VIVA-VOCE QUESTIONS:

1. Define transfer function.

2. What is DC servo motor? State the main parts.

3.

What is servo mechanism?4. Is this a closed loop or open loop system .Explain.

5. What is back EMF?6. What are the main parts of a DC servo motor?

7. Name the two types of servo motor.8. State the advantages and disadvantages of a DC servo motor.9. Give the applications of DC servomotor.

10. What is servo mechanism?11. What do you mean by field controlled DC servo motor?

Expt. No: Date:

DETERMINATION OF TRANSFER FUNCTION OF

AC SERVO MOTOR

AIM:To derive the transfer function of the given AC Servomotor.

APPARATUS / INSTRUMENTS REQUIRED:

S. No Description Range Type Quantity

1. AC servo motor trainer kit - 1

2. AC servo motor 1

3. Ammeter(0-1) A MC 1

(0-100) mA MI 1

4. Voltmeter(0 – 300) V MC 1

(0 – 75) V MI 1

5. Patch cords - As required

THEORY:

An AC servo motor is basically a two phase induction motor with some special designfeatures. The stator consists of two pole pairs (A-B and C-D) mounted on the inner periphery

of the stator, such that their axes are at an angle of 90 o in space. Each pole pair carries awinding, one winding is called reference winding and other is called a control winding. Theexciting current in the winding should have a phase displacement of 90o. The supply used to

drive the motor is single phase and so a phase advancing capacitor is connected to one of the phase to produce a phase difference of 90o.The rotor construction is usually squirrel cage or

drag-cup type. The rotor bars are placed on the slots and short-circuited at both ends by end





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rings. The diameter of the rotor is kept small in order to reduce inertia and to obtain goodaccelerating characteristics. The drag cup construction is employed for very low inertia

applications. In this type of construction the rotor will be in the form of hollow cylindermade of aluminium. The aluminium cylinder itself acts as short-circuited rotor conductors.

Electrically both the types of rotor are identical.

WORKING PRINCIPLE :

The stator windings are excited by voltages of equal magnitude and 90o phase difference.These results in exciting currents i1 and i2 that are phase displaced by 90o and have equal

values. These currents give rise to a rotating magnetic field of constant magnitude. Thedirection of rotation depends on the phase relationship of the two currents (or voltages). This

rotating magnetic field sweeps over the rotor conductors. The rotor conductor experience achange in flux and so voltages are induced rotor conductors. This voltage circulates currentsin the short-circuited rotor conductors and currents create rotor flux. Due to the interaction of

stator & rotor flux, a mechanical force (or torque) is developed on the rotor and so the rotorstarts moving in the same direction as that of rotating magnetic field.

GENERAL SCHEMATIC OF AC SERVOMOTOR:

FORMULAE USED:

Transfer function, Gm (s) = K m / (1+ sm)

Where

Motor gain constant, K m = K / FO + F

K is T / C

FO is T / N

Torque, T is 9.81 X R (S1 S2)

R is radius of the rotor in m

Frictional co-efficient, F = W / (2 N / 60)2





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Frictional loss, W is 30 % of constant loss in WattsConstant loss in watts = No load input – Copper loss

No load i/p = V (IR +IC)V is supply voltage, V

IR is current through reference winding, A

IC is current through control winding, ACopper loss in watts = IC2 R C

R C = 174 N is rated speed in rpm

Motor time constant, m = J / FO + F

Moment of inertia J is d4 L R ρ / 32

d is diameter of the rotor in m ( Given d =39.5 mm)

LR is length of the rotor in m (Given L R =76 mm)ρ is density = 7.8 X 102 gm / m

PROCEDURE:

1. DETERMINATION OF FRICTIONAL CO-EFFICIENT, F

1. Check whether the MCB is in OFF position.

2. Patch the circuit using the patching diagram.3. Switch ON the MCB4. Vary the control pot to apply rated supply voltage

5. Note the control winding current, reference winding current, supply voltage and

speed.6. Find the frictional co-efficient using the above values

OBSERVATIONS:

S. No.

Supply Voltage

V

(V)

Control winding

Current Ic

(A)

Reference Winding

Current Ir

(A)

Speed

N

(rpm)

CALCULATIONS:





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DETERMINATION OF TRANSFER FUNCTION OF AC SERVO MOTOR

PATCHING DIAGRAM TO DETERMINE FRICTIONAL CO-EFFICIENT F:





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PROCEDURE:

2. To determine the motor gain constant K m

DETERMINATION OF FO FROM TORQUE - SPEED CHARACTERISTICS:

1.

Check whether the MCB is in OFF position.2. Patch the circuit using the patching diagram.

3. Set the control pot in minimum position.4. Check whether the motor is in no load condition

5. Switch ON the MCB6. Vary the control pot and apply rated voltage to the reference phase winding and

control phase winding. Note down the no load speed.

7. Apply load in steps. For each load applied note down the speed and spring balancereadings. ( Take 3 or 4 sets of readings)

8. Reduce the load fully and allow the motor to run at rated speed.9. Repeat steps 7 and 8 for 75 % control winding voltage.

10. Draw the graph between speed and torque, the slope of the graph gives FO.

OBSERVATIONS:

S. No

Control voltage Vc1 = Control voltage Vc2 =

SpeedN

(rpm)

Spring Balancevalues Torque

T

(Nm)

SpeedN

(rpm)

Spring Balancevalues

TorqueT

(Nm)

S1

(kg)

S2

(kg)

S1

(kg)

S2

(kg)

MODEL GRAPH: TORQUE - SPEED CHARACTERISTICS





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DETERMINATION OF K FROM TORQUE - CONTROL VOLTAGE

CHARACTERISTICS:1. Check whether the MCB is in OFF position.

2. Patch the circuit using the patching diagram.

3.

Set the control pot in minimum position.4. Check whether the motor is in no load condition

5. Switch ON the MCB6. Vary the control pot and apply rated voltage to the reference phase winding and

control phase winding. Note down the no load speed.7. Load the motor gradually; the speed of the motor will decrease. Vary the control pot

and increase the control winding voltage till the speed obtained at no load is

reached. Note down control voltage and spring balance readings.8. Repeat step 7 for various speeds and tabulate. (for 1000 rpm)

9. Plot the graph between torque and control winding voltage. The slope of the graphgives the value of K.

OBSERVATIONS:

S. No

Speed N1 = Speed N2 =

Control

VoltageVc

(V)

Spring Balance

values

Torque

T

Nm

Speed

rpm

Spring Balance

values

Control

VoltageVc

VS1

(kgS2

kgS1

K g S2

K g

MODEL GRAPH: TORQUE - CONTROL VOLTAGE CHARACTERISTICS





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DETERMINATION OF TRANSFER FUNCTION OF AC SERVO MOTOR

PATCHING DIAGRAM TO DETERMINE MOTOR GAIN CONSTANT K M:





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CALCULATIONS:

RESULT:

The transfer function of AC servomotor is determined as


1. What are the main parts of an AC servomotor?

2. State the advantages and disadvantages of an AC servo motor.3. Give the applications of AC servomotor.

4. What do you mean by servo mechanism?5. What are the characteristics of an AC servomotor?





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Expt. No: Date:

DETERMINATION OF TRANSFER FUNCTION OFSEPARATELY EXCITED DC GENERATOR

AIM:

To obtain the transfer function of separately excited DC generator on no load andloaded condition.



THEORY:

Derivation of transfer function of separately excited DC generator is as follows,

Applying KVL to the field side,

ef = R f if + Lf (dif / dt) … (1)

Applying KVL to the armature side,

eg = R a ia + La (dia / dt) + R L ia … (2)

VL = R L ia … (3)

Also since eg α if , let eg = K g if … (4)

Taking Laplace transform of equation (1) we get

Ef (s) = R f If (s) + sLf If (s)

Ef (s) = If (s) [R f + sLf ]

If (s) = Ef (s) / [R f + sLf ] … (5)





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Taking Laplace transform of equation (2) we get

Eg (s) = R a Ia(s) + sLa Ia(s) + R L Ia(s)Eg (s) = Ia(s) [R a + sLa + R L] … (6)

Taking Laplace transform of equations (3) and (4) we getVL(s) = R L Ia( s)

Therefore, Ia( s) = VL(s) / R L … (7) Eg(s) = K g If (s) … (8)

Substituting. equations (7) and (8) in equation (6) we getK g If (s) = [R a + sLa + R L] [VL(s) / R L] … (9)

Substituting the value of If (s) in the above equation we get

K g Ef (s) / [R f + sLf ] = [R a + sLa + R L] [ VL(s) / R L]

Hence transfer function,VL(s) / Ef (s) = K g R L / [R f + sLf ] [R a + sLa + R L] …(10)

For unloaded condition, Ia = 0Therefore transfer function VL(s) / Ef (s) = K g / [R f + sLf ] … (11)

For loaded conditionLf = √ (Zf

2 – R f 2) / 2πf

La = √ (Za2 – R a

2) / 2πf

Transfer function VL(s) / Ef (s) = K g R L / [R f (R a + R L) (1+sτf ) (1 + sτa)] … (12)

where τf = Lf / R f and τa = La / (R a + R L)

FORMULAE USED:

Transfer function of DC generator,On no load condition: VL(s) / Ef (s) = K g / [R f + sLf ]

where K g is gain constantR f is field resistance in OhmsLf is field inductance in Henry

On loaded condition: VL(s) / Ef (s) = K g R L / [R f (R a + R L) (1+sτf ) (1 + sτa)]

where K g is gain constant

Field time constant τf = Lf / R f

R f is field resistance in Ohms

Lf is field inductance in Henry

Armature time constant τa = La / (R a + R L)

R a is armature resistance in OhmsLa is armature inductance in Henry





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PROCEDURE:

1. To determine the gain constant K g :

No load or open circuit characteristics:

1. Connections are made as shown in the circuit diagram

2. The motor field rheostat should be in minimum resistance position and the generatorfield rheostat should be in maximum resistance position or minimum potential

position while switching ON and switching OFF the supply side DPST switch.3. Ensure that the DPST switch on the load side is open.4. Switch ON the supply DPST switch.

5. Using the 3- point starter the DC motor is started and it is brought to rated speed byadjusting the motor field rheostat.

6. Keeping the DPST switch on the load side open, the generated voltage Eg and field

current If of generator is noted down by varying the generator field rheostat.

7.

The above step is repeated till 125 % of rated voltage is reached.8. A graph is plotted between Eg and If taking If along x- axis. A tangent to the linear

portion of the curve is drawn from the origin and slope of this line gives K g.

OBSERVATIONS:

MODEL GRAPH:

S. No.Field current, If

(A)

Induced Voltage, Eg

(V)





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CIRCUIT DIAGRAM:

To determine gain constant, K g:

CALCULATIONS:





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Load characteristics:

1. Connections are made as shown in the circuit diagram

2.

The motor field rheostat should be in minimum resistance position and the generatorfield rheostat should be in maximum resistance position or minimum potential

position while switching ON and switching OFF the supply side DPST switch.3. Ensure that the DPST switch on the load side is open.

4. Switch ON the supply DPST switch5. The generator is brought to its rated voltage by varying the generator field rheostat.6. The DPST switch on the load side is closed, and the load is varied for convenient

steps of load current up to 120 % of its rated capacity and the voltmeter V L andammeter Ia readings are observed. On each loading the speed should be maintained at

rated speed.7. A graph is plotted between VL and IL taking IL on x- axis. The slope of the graph

gives K g.

OBSERVATIONS:

MODEL GRAPH:

S. No.Terminal Voltage, VL

(V)

Load Current, IL

(A)





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PROCEDURE:

2. To determine field Inductance Lf :

1. Connections are made as per the circuit diagram.2. Auto transformer is varied in steps for different voltages and corresponding voltmeter

and ammeter readings are noted down.3. Field impedance Zf is calculated as V/I and the average value of Zf is obtained.4. Field resistance (R f ) is measured using multimeter.

5. Field inductance (Lf ) can be calculated using formulaLf = √ (Zf

2 – R f 2) / 2πf

CIRCUIT DIAGRAM:

OBSERVATIONS: S. No

Field Voltage, V(V)

Field Current, I

(A)Field Impedence, Zf

(Ohms)

CALCULATIONS:





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PROCEDURE:

3. Determination of armature inductance La :

1. Connections are made as per the circuit diagram.2. Auto transformer is varied in steps for different voltages and corresponding voltmeter

and ammeter readings are noted down.3. Armature impedance Za is calculated as V/I and the average value of Za is obtained.4. Armature resistance R a is measured using multimeter.

5. Armature inductance La can be calculated using formula,La = √ (Za

2 – R a2) / 2πf

CIRCUIT DIAGRAM:

OBSERVATIONS:

S. NoArmature

Voltage, V (V)

Armature

Current, I

(A)

Armature Impedence, Za

(Ohms)

CALCULATIONS:





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CALCULATIONS:

RESULT:

The transfer function of separately excited DC generator is determined as





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Expt. No: Date:

DETERMINATION OF TRANSFER FUNCTION OF DC MOTOR

AIM:

To obtain the transfer function of field controlled DC motor.



THEORY:





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FIELD CONTROLLED MOTOR:

FORMULAE USED:

Transfer function of field controlled DC motor,

(s) / Vf (s) = K m / [s (1+sτf ) (1 + sτm)]

where

Motor gain constant, K m =K tf / (BR f )K tf is motor torque constant

Torque, T is 9.81 X R (S1 S2)

R is radius of the brake drum in mR = circumference of the brake drum/ (2 П) B is viscous co-efficient of friction


Field time constant τf = Lf / R f


Lf is field inductance in HenryLf = √ (Zf

2 – R f 2) / 2πf

Zf is field impedence in Ohms


Mechanical time constant τm = J/B

Moment of inertia J = Pav / [2П2(N12 – N2

2 )((1/t1)-(1/t2))]

Average power delivered to the load, Pav= (V1I1 + I12 R a + V2I2 + I22 R a) / 2R a is armature resistance in OhmsLa is armature inductance in Henry

t2 is time taken on load in secst1 is time taken on no load in secs

Viscous Co-efficient of friction, B = Pstray / (N1 + N2)2

Stray loss, Pstray = [2П2(N12 – N2

2 )] J / t1





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PROCEDURE:

1. To determine motor torque constant, K tf :

1.

Connections are made as shown in the circuit diagram2. The armature current Ia of the motor is set to some value by adjusting the armature

circuit resistance. This value of Ia is maintained constant throughout the experiment.3. The field current If is varied in steps by adjusting the field rheostat and for each value

of If the brake drum is adjusted such that it just fails to rotate. The correspondingreadings of ammeter and spring balances are noted.

4. The value of torque for each value of If is calculated

5. A graph is plotted between torque T and field current IF taking IF along x-axis. Theslope of the graph gives the value of K tf

OBSERVATIONS:

S. No.

Armature current

Ia

(A)

Field current

If

(A)

Spring balance readings Torque

T

(Nm)

S1

(kg)

S2

(kg)

MODEL GRAPH:





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CIRCUIT DIAGRAM:

CALCULATIONS:





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PROCEDURE

2. To determine field Inductance Lf :

1. Connections are made as per the circuit diagram.

2.

Auto transformer is varied in steps for different voltages and corresponding voltmeterand ammeter readings are noted down.

3. Field impedance Zf is calculated as V/I and the average value of Zf is obtained.4. Field resistance (R f ) is measured using multimeter.

5. Field inductance (Lf ) can be calculated using formulaLf = √ (Zf

2 – R f 2) / 2πf

CIRCUIT DIAGRAM:

OBSERVATIONS: S. No.

Field Voltage, V

(V)

Field Current, I

(A)

Field Impedence, Zf

(Ω)

CALCULATIONS:





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PROCEDURE:

3. To determine moment of inertia J and Viscous friction Co-efficient B:

1.

Connections are made as shown in the circuit diagram2. The field current of the motor is set to some value by adjusting the field resistance.

3. DPDT switch is thrown to position (1,11) and the motor is made to run at a speed N 1

(1700 rpm) by adjusting the armature rheostat.

4. DPDT switch is opened from position (1,11) and the stop watch is startedsimultaneously. The time taken t1 for the speed to drop from N1(1700 rpm) to N2

( 1300 rpm) is noted.

5. Again the DPDT switch is thrown to position (1,11) and the motor is made to run at aspeed greater than N1 (1700 rpm) by adjusting the armature rheostat.

6. DPDT switch is thrown to position (2,21) and the stop watch is started when the motorspeed reaches N1 (1700 rpm). The time taken t2 for the speed to drop from N1 (1700

rpm) to N2( 1300 rpm) is noted. Simultaneously the readings of the ammeter andvoltmeter corresponding to N1 and N2 are noted.

OBSERVATIONS:

S. No. N1

(rpm)

t1

(Sec)

V1

(V)

I1

(A)

N2

(rpm)

T2

(Sec)

V2

(V)

I2

(A)

CALCULATIONS:





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CIRCUIT DIAGRAM:

CALCULATIONS:





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CALCULATIONS:

RESULT:

The transfer function of field controlled DC motor is determined as





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Expt. No: Date:

DC POSITION CONTROL SYSTEM

AIM:

To study the characteristics of a DC position control system.


i) DC position control kit and Motor unit

ii) Multimeter

THEORY:

A DC position control system is a closed loop control system in which the position of themechanical load is controlled with the position of the reference shaft. A pair of

potentiometers acts as error-measuring device. They convert the input and output positionsinto proportional electric signals. The desired position is set on the input potentiometer and

the actual position is fed to feedback potentiometer. The difference between the two angular positions generates an error signal, which is amplified and fed to armature circuit of the DCmotor. The tachogenerator attached to the motor shaft produces a voltage proportional to the

speed which is used for feedback. If an error exists, the motor develops a torque to rotate theoutput in such a way as to reduce the error to zero. The rotation of the motor stops when the

error signal is zero, i.e., when the desired position is reached.

PROCEDURE:

1. The input or reference potentiometer is adjusted nearer to zero initially(R ).

2. The command switch is kept in continuous mode and some value of forward gain K Ais selected.

3. For various positions of input potentiometer (R ) the positions of the response

potentiometer (0) is noted. Simultaneously the reference voltage (VR ) measured

between the terminals VR & E and the output voltage (VO) measured between theterminals VO & E are noted.

4. A graph is plotted with 0 along y-axis and R along x-axis.





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OBSERVATIONS:

S. No

Reference

angular position,

θR

(degrees)

Output angular

position, θO

(degrees)

Reference

Voltage, Vr

(V)

Output

VoltageVO

(V)

K A = K A = K A = K A = K A = K A = K A = K A =

MODEL GRAPH:





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DC POSITION CONTROL SYSTEM





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RESULT:

The DC position control system characteristics are studied and corresponding graphs are drawn.





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Expt. No. Date:

ANALOG SIMULATION OF TYPE – 0 and TYPE – 1 SYSTEMS

AIM:

To study the time response of first and second order type –

0 and type- 1 systems.


1. Linear system simulator kit

2. CRO3. Patch cords

FORMULAE USED:

Damping ratio, = (ln MP)2 / (2 +(ln MP)2)Where MP is peak percent overshoot obtained from the time response graph

Undamped natural frequency, n = / [t p (1 - 2)]

where t p is the peak time obtained from the time response graph

Closed loop transfer function of the type – 0 second order system is

C(s)/R(s) = G(s) / [1 + G(s) H(s)]

whereH(s) = 1G(s) = K K 2 K 3 / (1+sT1) (1 + sT2)

where K is the gain

K 2 is the gain of the time constant –

1 block =10K 3 is the gain of the time constant – 2 block =10T1 is the time constant of time constant – 1 block = 1 msT2 is the time constant of time constant – 2 block = 1 ms

Closed loop transfer function of the type – 1-second order system is

C(s)/R(s) = G(s) / [1 + G(s) H(s)]where

H(s) = 1

G(s) = K K 1 K 2 / s (1 + sT1)where K is the gain

K 1 is the gain of Integrator = 9.6K 2 is the gain of the time constant – 1 block =10T1 is the time constant of time constant – 1 block = 1 ms

THEORY:

The type number of the system is obtained from the number of poles located at origin in a givensystem. Type – 0 system means there is no pole at origin. Type – 1 system means there is one

pole located at the origin. The order of the system is obtained from the highest power of s in thedenominator of closed loop transfer function of the system. The first order system is





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characterized by one pole or a zero. Examples of first order systems are a pure integrator and asingle time constant having transfer function of the form K/s and K/(sT+1). The second order

system is characterized by two poles and up to two zeros. The standard form of a second order

system is G(s) = n2

/ (s2 + 2ns + n

2) where is damping ratio and n is undamped natural

frequency.

PROCEDURE:

1. To find the steady state error of type – 0 first order system

1. Connections are made in the simulator kit as shown in the block diagram.

2. The input square wave is set to 2 Vpp in the CRO and this is applied to the REFterminal of error detector block. The input is also connected to the X- channel of CRO.

3. The output from the simulator kit is connected to the Y- channel of CRO.

4. The CRO is kept in X-Y mode and the steady state error is obtained as the verticaldisplacement between the two curves.

5. The gain K is varied and different values of steady state errors are noted.

Block diagram of Type-0 first order system

OBSERVATIONS:

S. No. Gain, K Steady state error, ess

1

2

3





39

TRACES FROM CRO:

For Gain, K =

For Gain, K =

For Gain, K=





40

LINEAR SYSTEM SIMULATOR

PATCHING DIAGRAM TO OBTAIN THE STEADY STATE ERROR OF TYPE – 0 FIRST ORDER SYSTEM





41

2. To find the steady state error of type – 1 first order system

1. The blocks are Connected using the patch chords in the simulator kit.

2. The input triangular wave is set to 2 Vpp in the CRO and this applied o the REF

terminal of error detector block. The input is also connected to the X- channel of CRO.3. The output from the system is connected to the Y- channel of CRO.

4. The experiment should be conducted at the lowest frequency to allow enoughtime for the step response to reach near steady state.

5. The CRO is kept in X-Y mode and the steady state error is obtained as the verticaldisplacement between the two curves.

6. The gain K is varied and different values of steady state errors are noted.

7. The steady state error is also calculated theoretically and the two values are compared.

Block diagram of Type- 1 First order system

OBSERVATIONS:

S. No. Gain, K Steady state error, ess

1

2

3





42

TRACES FROM CRO:

For Gain, K =

For Gain, K =

For Gain, K =





43

LINEAR SYSTEM SIMULATOR PATCHING DIAGRAM TO OBTAIN THE STEADY STATE ERROR OF TYPE – 1 FIRST ORDER SYSTEM





44

3. To find the closed loop response of type – 0 and type- 1 second order system

1. The blocks are connected using the patch chords in the simulator kit.2. The input square wave is set to 2 Vpp in the CRO and this applied to the REF terminal

of error detector block. The input is also connected to the X- channel of CRO.

3. The output from the system is connected to the Y- channel of CRO.4. The output waveform is obtained in the CRO and it is traced on a graph sheet. From

the waveform the peak percent overshoot, settling time,rise time, peak time are

measured. Using these values n and are calculated.

5. The above procedure is repeated for different values of gain K and the values arecompared with the theoretical values.

Block diagram to obtain closed loop response of Type-0 second order system

OBSERVATIONS:

S. No.Gain

K

Peak

percentOvershoot

%MP

Rise

timetr

(sec)

Peak

Timetp

(sec)

Settling

timets

(sec)

Damping

ratio

UndampedNatural

frequency

n

(rad/sec)

1

2

TRACES FROM CRO:

For Gain, K = For Gain, K =





45

Block diagram to obtain closed loop response of Type-1 second order system

OBSERVATIONS:

S. No.Gain

K

Peak

percent

Overshoot

%MP

Rise

time

tr

(sec)

Peak

Time

tp

(sec)

Settling

time

ts

(sec)

Dampingratio

Undamped

Naturalfrequency

n

(rad/sec)

1

2

TRACES FROM CRO:

For Gain, K = For Gain, K =





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LINEAR SYSTEM SIMULATOR PATCHING DIAGRAM TO OBTAIN THE CLOSED LOOP RESPONSE OF TYPE – 0 SECOND ORDER SYSTEM





47

LINEAR SYSTEM SIMULATOR

PATCHING DIAGRAM TO OBTAIN THE CLOSED LOOP RESPONSE OF TYPE – 1 SECOND ORDER SYSTEM





48

CALCULATIONS:

RESULT:

The time response of first and second order type-0 and type-1 systems are studied.


1. Define order and type number.2. What are dominant poles?3. What is a closed loop system?

4. What is the effect of negative feedback?5. What are poles and zeros of a system?

6. Define transfer function.





49

Expt. No. Date:

DIGITAL SIMULATION OF FIRST ORDER SYSTEMS

AIM:

To digitally simulate the time response characteristics of a linear system without

non- linearities and to verify it manually.

APPARATUS REQUIRED:

A PC with MATLAB package

THEORY:

The time response characteristics of control systems are specified in terms of time

domain specifications. Systems with energy storage elements cannot respondinstantaneously and will exhibit transient responses, whenever they are subjected to inputs or

disturbances.The desired performance characteristics of a system of any order may be specified in

terms of transient response to a unit step input signal. The transient response characteristics

of a control system to a unit step input is specified in terms of the following time domainspecifications

Delay time td Rise time tr Peak time t p

Maximum peak overshoot M p Settling time ts

STUDY OF BASIC MATLAB COMMANDS:

The name MATLAB stands for MATRIX LABORATORY. MATLAB was originallywritten to provide easy access to matrix software developed by the LINPACK and EISPACK

projects. Today, MATLAB engines incorporate the LAPACK and BLAS libraries,embedding the state of the art in software for matrix computation. It has evolved over a period of years with input from many users. In university environments, it is the standard

instructional tool for introductory and advanced courses in MATHEMATICS,

ENGINEERING, AND SCIENCE. In industry, MATLAB is the tool of choice for high-

productivity research, development, and analysis.

MATLAB is a high-performance language for technical computing. It integratescomputation, visualization, and programming in an easy-to-use environment where problemsand solutions are expressed in familiar mathematical notation. Typical uses include,

Math and computation Algorithm development

Data acquisition Modeling, simulation, and prototyping Data analysis, exploration, and visualization

Scientific and engineering graphics





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Application development, including graphical user interface building

It is an interactive system whose basic data element is an array that does not requiredimensioning. This allows you to solve many technical computing problems, especially those

with matrix and vector formulations, in a fraction of the time it would take to write a program

in a scalar non-interactive language such as C or Fortran. It also features a family of add-onapplication-specific solutions called toolboxes. Very important to most users of MATLAB,

toolboxes allow you to learn and apply specialized technology. Toolboxes are comprehensivecollections of MATLAB functions (M-files) that extend the MATLAB environment to solve

particular classes of problems. Areas in which toolboxes are available include SIGNAL

PROCESSING, CONTROL SYSTEMS, NEURAL NETWORKS, FUZZY LOGIC,

WAVELETS, SIMULATION, AND MANY OTHERS.

Some practical examples of first order systems are RL and RC circuits.

PROCEDURE:

1. Derive the transfer function of a RL series circuit.

2. Assume R= 1 Ohms L = 0. 1 H. Find the step response theoretically and plot it on agraph sheet.

3. To build a SIMULINK model to obtain step response / sine response of a first order

system, the following procedure is followed :1. In MATLAB software open a new model in SIMULINK library browser.

2. From the continuous block in the library drag the transfer function block.3. From the source block in the library drag the step input/ sine input.4. From the sink block in the library drag the scope.

5. From the math operations block in the library drag the summing point.6. Connect all to form a system and give unity feedback to the system.

7. For changing the parameters of the blocks connected double click therespective block.

8. Start simulation and observe the results in scope. (Use a mux from the signal

routing block to view more than one graph in the scope)9. Compare the simulated and theoretical results.

BLOCK DIAGRAM:

Step response of a fi rst order system:





51

Sine response of a fi rst order system:

2. MATLAB (m-file) program to obtain the step response and impulse response

% MATLAB program to find the step response

num=[ ];

den=[ ];sys = tf (num,den);

step (sys);grid

OUTPUT: (Paste the graph obtained from PC)

% MATLAB program to find the impulse response

num=[ ];den=[ ];

sys = tf (num,den);impulse (sys);

grid





52


CALCULATIONS:

Unit step response of the given RL series circuit:





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Unit Impulse response of the given RLC series circuit:

RESULT:

The time response characteristics of a first order system is simulated digitally and verified

manually.


1. What is MATLAB?2. What is the use of MATLAB Package?

3. What are the toolboxes available in MATLAB?4. What is the use of a simulation?5. Differentiate real time systems and simulated systems.

6. Give two examples for first order system.7. Name the standard test signals used in control system.

8. What is time response?





54

Expt. No: Date:

DIGITAL SIMULATION OF SECOND ORDER SYSTEMS

AIM:

To digitally simulate the time response characteristics of a second order system and verifymanually.

APPARATUS REQUIRED

A PC with MATLAB Software

THEORY

The time characteristics of control systems are specified in terms of time domainspecifications. Systems with energy storage elements cannot respond instantaneously and

will exhibit transient responses, whenever they are subjected to inputs or disturbances. Thedesired performance characteristics of a system of any order may be specified in terms oftransient response to a unit step input signal. The transient response characteristics of a

control system to a unit step input is specified in terms of the following time domainspecifications:

Delay time td Rise time tr Peak time t p

Maximum overshoot M p Settling time ts

PROCEDURE:

1. Derive the transfer function of a RLC series circuit.2. Assume R= 1 Ohms, L = 0. 1 H and C = 1 micro Farad. Find the step response

theoretically and plot it on a graph sheet.3. To build a SIMULINK model to obtain step response / sine response of a second

order system, the following procedure is followed:

1. In MATLAB software open a new model in SIMULINK library browser.2. From the continuous block in the library drag the transfer function block.

3. From the source block in the library drag the step input/ sine input.4. From the sink block in the library drag the scope.

5. From the math operations block in the library drag the summing point.6. Connect all to form a system and give unity feedback to the system.7. For changing the parameters of the blocks connected double click the

respective block.8. Start simulation and observe the results in scope. (Use a mux from the signal

routing block to view more than one graph in the scope)

9. From the step response obtained note down the rise time, peak time, peakovershoot and settling time.

10. Compare the simulated and theoretical results.





55

BLOCK DIAGRAM:

Step response of a second order system:

Sine response of a second order system:

2. MATLAB program to obtain the step response and impulse response of second order

system.

% MATLAB program to find the step responsenum=[ ];den=[ ];

sys = tf (num,den);step (sys);






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% MATLAB program to find the impulse response

num=[ ];

den=[ ];sys = tf (num,den);

impulse (sys);


CALCULATIONS:

Unit step response of the given RLC series circuit:





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Unit impulse response of the given RLC series circuit:





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RESULT:

The time response characteristics of the given second order system is simulated digitally andverified manually.


1. What is MATLAB?2. What is the use of MATLAB Package?

3. What are the toolboxes available in MATLAB?4. What is the use of a simulation?

5. Differentiate real time systems and simulated systems.6. Give two examples for second order system.7. Name the standard test signals used in control system.

8. What is time response?





59

Expt. No: Date:

STABILITY ANALYSIS OF LINEAR SYSTEMS

a. USING BODE PLOT

AIM:

To obtain the bode plot and check for stability of the system with open loop transfer function,G(S) =

APPARATUS REQUIRED:


THEORY:

A Linear Time-Invariant Systems is stable if the following two notions of system stability are

satisfied When the system is excited by Bounded input, the output is also a Bounded

output.

In the absence of the input, the output tends towards zero, irrespective of theinitial conditions.

The following observations are general considerations regarding system stability,

If all the roots of the characteristic equation have negative real parts, then theimpulse response is bounded and eventually decreases to zero, then system is

stable. If any root of the characteristic equation has a positive real part, then system is

unstable.

If the characteristic equation has repeated roots on the jω -axis, then system isunstable.

If one are more non-repeated roots of the characteristic equation on the jω -axis, then system is unstable.

BODE PLOT :

Consider a Single-Input Single-Output system with transfer function

C(s) b0 sm + b1 sm-1 + ……+ bm

=R(s) a0 sn + a1sn-1 + ……+an

Where m < n.

Rule 1 A system is stable if the phase lag is less than 180˚ at the frequency

for which the gain is unity (one).





60

Rule 2 A system is stable if the gain is less than one (unity) at the frequencyfor which the phase lag is 180˚.

The application of these rules to an actual process requires evaluation of the gain and phase

shift of the system for all frequencies to see if rules 1 and 2 are satisfied. This is obtained by

plotting the gain and phase versus frequency. This plot is called BODE PLOT. The gainobtained here is open loop gain. The exact terminology is in terms of a Gain Margin and

Phase Margin from the limiting values quoted.

If the phase lag is less than 140˚ at the unity gain frequency, the system isstable. This then, is a 40˚ Phase Margin from the limiting values of 180˚.

If the gain is 5dB below unity (or a gain of about 0.56) when the phase lag is180˚, the system is stable. This is 5dB Gain Margin.

PROCEDURE:

Step 1: Write a program to obtain the Bode plot for the given system.Step 2: Assess the stability of given system using the plot obtained.

PROGRAM

%BODE PLOT OF THE SYSTEM

%Enter the numerator and denominator of the transfer function

num=[ ];

den=[ ];

sys=tf(num,den)

%Specify the frequency range and enter the command

w=logspace(-2,4,1000);

bode(sys,w)

xlabel('Frequency')

ylabel( ' Phase angle in degrees Magnitude of G(s) in decibels')

title('Bode Plot of the system ')

%To determine the Gain Margin, Phase Margin, Gain crossover frequency and%Phase cross over frequency

margin(sys)

[ Gm, Pm, Wpc, Wgc ]= margin (sys)





61

MANUAL CALCULATIONS:





62





63

OUTPUT (from manual calculation):

OUTPUT (from program):

RESULT:

The Bode plot is drawn for the given transfer function using MATLAB and verifiedmanually. From the plot obtained, the system is found to be ______________.


1. Define stability of Linear Time Invariant System.2. Give the stability conditions of system using Pole-Zero plot.

3. Define Bode Plot.4. What is the use of Bode Plot?

5. What the conditions of stability are in Bode plot?6. Define Stability criteria.7. Define Limits of stability.

8. Define safe regions in stability criteria.9. Define Phase margin and Gain margin.





64

b. Using Root Locus

AIM:

To obtain the Root locus plot and to verify the stability of the system with transfer function,

G(s) =

APPARATUS REQUIRED:


THEORY:

ROOT LOCUS PLOT:

The characteristic of the transient response of a closed-loop system is related to the location

of the closed loop poles. If the system has a variable loop gain, then the location of theclosed-loop poles depend on the value of the loop gain chosen. A simple technique known as“Root Locus Technique” used for studying linear control systems in the investigation of the

trajectories of the roots of the characteristic equation.

This technique provides a graphical method of plotting the locus of the roots in the s-plane as

a given system parameter is varied over the complete range of values (may be from zero toinfinity). The roots corresponding to a particular value of the system parameter can then be

located on the locus or the value of the parameter for a desired root location can bedetermined form the locus. The root locus is a powerful technique as it brings into focus thecomplete dynamic response of the system. The root locus also provides a measure of

sensitivity of roots to the variation in the parameter being considered. This technique isapplicable to both single as well as multiple-loop systems.

PROCEDURE:

1. Write a program to obtain the root locus plot for the given system.2. Assess the stability of given system using the plot obtained.

PROGRAM:

%ROOT LOCUS OF THE SYSTEM%

num=[ ]

den=[ ]

sys=tf(num,den)

rlocus(sys)

v=[-10,10,-8,8];

axis(v)





65

xlabel('Real Axis')

ylabel('Imaginary Axis')

title('Root Locus of the system')

title('Root Locus Plot of the system ')






66





67

OUTPUT (from manual calculation)

OUTPUT (from program):

RESULT:

The Root locus plot is drawn for the given transfer function, G(s)= ___________________

using MATLAB and the range of gain K for stability is______________.


1. Define root locus technique.2. What are the conditions of stability in root locus criteria?3. What is the advantage of root locus technique?

4. Which method of stability analysis is more advantageous?5. How the stability of unstable is improved?

6. What are the methods to improve the stability?7. What is the use of compensators?8. What do you mean by Root-Loci?

9. What is complementary Root Loci?10. What are contours?

11. State the basic properties of Root Locus.12. How would you find the number of branches of Root Loci?13. How are the break away points of the root locus determined?

14. How is the point of intersection of the asymptotes with real axis found out.





68

c. USING NYQUI ST PLOT

AIM:

To obtain the Nyquist plot and check the stability of the system using Nyquist Stability

Criterion for the given unity feedback system with transfer function

G(s)H(s) =

APPARATUS REQUIRED


THEORY:

NYQUIST STABILITY CRITERION:

POLAR PLOTS / NYQUIST PLOTS:

The sinusoidal transfer function G(jω) is a complex function is given by

G(jω) = Re[ G(jω)] + j Im[G(jω)] or

G(jω) = G(jω) ∟G(jω) = M ∟Φ -----------(1)

From equation (1), it is seen that G(jω) may be represented as a phasor of magnitude M and phase angle Φ. As the input frequency varies from 0 to ∞, the magnitude M and phase angleΦ changes and hence the tip of the phasor G(jω) traces a locus in the complex plane. Thelocus thus obtained is known as POLAR PLOT. The major advantage of the polar plot lies

in stability study of systems. Nyquist related the stability of a system to the form of these plots. Polar plots are referred as NYQUIST PLOTS.

PROCEDURE:

1. Write a program to obtain the Nyquist plot for the given system.2. Assess the stability of given system using the plot obtained.





69

PROGRAM

%NYQUIST PLOT%Enter the numerator and denominator of the transfer function

num=[ ]

den=[ ]sys=tf(num,den)

%Specify the frequency range and enter the command

nyquist(sys)v=[ ]axis(v)

xlabel('Real Axis');ylabel('Imaginary Axis');

title('Nyquist Plot of the system „)






70

OUTPUT ( from Manual calculation)

OUTPUT (from program)





71

RESULT:

The Nyquist plot is drawn for the given transfer function, G(s) = ______________________

using MATLAB and the system is found to be ______________________.


1. What is polar plot?2. What is Nyquist plot?

3. Define the conditions of stability in polar plot.4. What is the use and advantage of polar plot?

5. State Nyquist stability criterion.





72

Expt. No: Date:

STEPPER MOTOR CONTROL SYSTEM





73



70638637 control system lab manual opt

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